6.4 L’Hôpital’s Rule

Suppose we want to analyze the behavior of the following function:

$f(x)=\frac{\ln x^2}{x^2-1}$

While $f$ is not defined at $x=-1$ and $1$, $f$ behaves certain ways near $x=-1$ and $1$. In order to understand the behavior of the function $f$, we can take the limit as $x$ approaches 1 and -1.

$\underset{x\to -1}{lim}\frac{\ln x^2}{x^2-1}$

$\underset{x\to 1}{lim}\frac{\ln x^2}{x^2-1}$

The issue here is that we cannot actually calculate the limit since both the numerator and denominator of the function go to 0 as they approach $x=-1$ or $1$.

$\underset{x\to -1}{lim}\frac{\ln x^2}{x^2-1}=\frac{0}{0}$

$\underset{x\to 1}{lim}\frac{\ln x^2}{x^2-1}=\frac{0}{0}$

How about evaluating the following limits?

$\underset{x\to -\mathrm{\infty }}{lim}\frac{\ln x^2}{x^2-1}$

$\underset{x\to \mathrm{\infty }}{lim}\frac{\ln x^2}{x^2-1}$

Again, we cannot find the limits, this time because both the numerator and the denominator of the function go to infinity.

$\underset{x\to -\mathrm{\infty }}{lim}\frac{\ln x^2}{x^2-1}=\frac{\mathrm{\infty }}{\mathrm{\infty }}$

$\underset{x\to \mathrm{\infty }}{lim}\frac{\ln x^2}{x^2-1}=\frac{\mathrm{\infty }}{\mathrm{\infty }}$

In order to find the limit in such cases, we can use L’Hôpital’s rule. Suppose $f$ and $g$ are differentiable on an open interval $I$ except possibly at a point $c$ contained in $I$. Moreover, suppose that

$\underset{x\to c}{lim}f(x)=0and\underset{x\to c}{lim}g(x)=0$

or that

$\underset{x\to c}{lim}f(x)=\pm \mathrm{\infty }and\underset{x\to c}{lim}g(x)=\pm \mathrm{\infty }$

Then,

$\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$

if the following limit exists:

$\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Basically, L’Hôpital’s rule says that if a limit of both the numerator and denominator go to 0 or if a limit of both the numerator and denominator go to positive or negative infinity, then we can take the derivative of the numerator and denominator separately and then take the limit, unless we can somehow eliminate some term that makes both the numerator and the denominator go to 0 together or to positive or negative infinity together.

What if the limit of the derivative of both the numerator and the denominator go to 0 together or infinity together? We can keep taking derivatives until such behavior stops. Here are some examples.

Example

Let’s start with the limit from above that we did not solve. Solve the following limit:

$\underset{x\to \mathrm{\infty }}{lim}\frac{\ln x^2}{x^2-1}$

As we know, both the numerator and the denominator go to infinity as $x$ approaches infinity. We apply L’Hôpital’s rule:

$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}\frac{\ln x^2}{x^2-1}& =\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{2x}{x^2}}{2x}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{2}{x}}{2x}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x^2}\\ & =\frac{1}{\mathrm{\infty }}\\ & =0\end{array}$

How about this one?

$\underset{x\to 1}{lim}\frac{\ln x^2}{x^2-1}$

Since both the numerator and the denominator will be 0 with the limits, we apply L’Hôpital’s rule again.

$\begin{array}{rl}\underset{x\to 1}{lim}\frac{\ln x^2}{x^2-1}& =\underset{x\to 1}{lim}\frac{\frac{2x}{x^2}}{2x}\\ & =\underset{x\to 1}{lim}\frac{\frac{2}{x}}{2x}\\ & =\underset{x\to 1}{lim}\frac{1}{x^2}\\ & =\frac{1}{1}\\ & =1\end{array}$

Let's look at another example. Find the limit of

$\underset{x\to \mathrm{\infty }}{lim}\frac{4x^3}{e^x}$

As we know, the limit of $4x^3$ with $x$ approaching infinity is infinity. Likewise, the limit of $e^x$ with $x$ approaching infinity is infinity. If we apply L’Hôpital’s rule, then we get

$\underset{x\to \mathrm{\infty }}{lim}\frac{4x^3}{e^x}=\underset{x\to \mathrm{\infty }}{lim}\frac{12x^2}{e^x}$

Since we cannot solve the problem yet, we will apply L’Hôpital’s rule until either or both the numerator and the denominator stop going to infinity.

$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}\frac{4x^3}{e^x}& =\underset{x\to \mathrm{\infty }}{lim}\frac{12x^2}{e^x}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{24x}{e^x}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{24}{e^x}\end{array}$

The limit of 24 with $x$ approaching infinity is 24, while the denominator still approaches infinity, so we get the following:

$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}\frac{4x^3}{e^x}& =\underset{x\to \mathrm{\infty }}{lim}\frac{12x^2}{e^x}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{24x}{e^x}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{24}{e^x}\\ & =\frac{24}{\mathrm{\infty }}\\ & =0\end{array}$

Understanding and being able to apply L’Hôspital’s rule will help you understand how each function behaves at important points of interest.